Q: 67^71^73^79/ 83 find the R?
@saurav205 said:combine the terms:21^3 + 27^3 and 23^3 + 25^3...use the a^3 + b^3 formula.....try this out...you should be able to get it...and please do not post the OA along with the question....
after combining do not calculate the whole thing..
(27+21)(27^2 + 21^2 -27*21) + (25+23)(25^2+23^2-25*23)
take 48 out ..
48(odd+odd-odd) + 48(odd+odd-odd)
48(even - odd) + 48(even -odd)
48(odd) + 48(odd)
48(odd + odd)
48(even)
so def by 96 as ek 2 toh hoga pakka se...
PS: OA jab pata hota hai toh aaise ideas lag jaate hain..thats why I said dont put the OA with the question....
@Abir1103 said:Find the remainder wen 21^3 + 23^3 + 25^3 + 27^3 is divided by 96 ?ans-> 0 . need soln plz
(a+b)(a^2+b^2-ab) ..
21^3+27^3 = 48(21^2 + 27^2 -21*27)
23^3+25^3=48 (23^2+25^2 -23*25)
48(1182)
this is divisible by 96.
answer has to be zero.
21^3+27^3 = 48(21^2 + 27^2 -21*27)
23^3+25^3=48 (23^2+25^2 -23*25)
48(1182)
this is divisible by 96.
answer has to be zero.
@Abir1103 said:Find the remainder wen 21^3 + 23^3 + 25^3 + 27^3 is divided by 96 ?ans-> 0 . need soln plz
96 = 32*3
(21^3 + 23^3 + 25^3 + 27^3)%32 = {(-11)^3 + (-9)^3 + (-7)^3 + (-5)^3}%32 = {-(1331+729+343+125)}%32 = 0
(21^3 + 23^3 + 25^3 + 27^3)%3 = (0+2^3+1+0)%3 = 0
so 0...
otherwise u cld also use (a^n+b^n) is div by (a+b) where n is odd...
i thnk (a^n+ b^n )/ (a+b) is the easiest method
@viewpt said:Q: 67^71^73^79/ 83 find the R?
e(83) is 82...
71^73^79%82
e(82) is 40..
73^79%40
e(40) is 16..
79%16 = 15
=> 33^15%40 = 17...
=> 11^17%82 = 69..
=> 67^69%83 = 43??
kya bekaar ques tha ye.. 
@Logrhythm said:96 = 32*3(21^3 + 23^3 + 25^3 + 27^3)%32 = {(-11)^3 + (-9)^3 + (-7)^3 + (-5)^3}%32 = {-(1331+729+343+125)}%32 = 0(21^3 + 23^3 + 25^3 + 27^3)%3 = (0+2^3+1+0)%3 = 0so 0...otherwise u cld also use (a^n+b^n) is div by (a+b) where n is odd...
last waala method toh strike hi nai kiya....
main toh even odd karta reh gaya...
@jain4444 said:The total number of Integral solutions of uvw²x²y² =277200
@scrabbler said:Getting 180....feel like I have missed something....retrying Edit: I give up. Too much noise in the street outside....some procession...can't concentrate regardsscrabbler
@ScareCrow28 said:Too many cases Confusion he confusion hai, solution ka pata nahi..
@iLoveTorres said:144?
its 46080
u = (2^a1)(3^b1)(5^c1)(7^d1)(11^d1)
v = (2^a2)(3^b2)(5^c2)(7^d2)(11^d2)
w = (2^a3)(3^b3)(5^c3)(7^d3)(11^d3)
x = (2^a4)(3^b4)(5^c4)(7^d4)(11^d4)
y = (2^a5)(3^b5)(5^c5)(7^d5)(11^d5)
a1 + a2 + 2a3 + 2a4 + 2a5 = 4
find the solutions of this equation, it will be 20
similarly for remaining 4 prime factors its 6, 6, 2, 2 ways
So, 20*6*6*2*2 positive integral solutions
Now, w, x, y can be positive or negative (as it doesn't matter), so 2*2*2 = 8 ways
For u and v, either both positive or both negative, so 2 ways
So, (20*6*6*2*2)*(8*2) integral solutions
v = (2^a2)(3^b2)(5^c2)(7^d2)(11^d2)
w = (2^a3)(3^b3)(5^c3)(7^d3)(11^d3)
x = (2^a4)(3^b4)(5^c4)(7^d4)(11^d4)
y = (2^a5)(3^b5)(5^c5)(7^d5)(11^d5)
a1 + a2 + 2a3 + 2a4 + 2a5 = 4
find the solutions of this equation, it will be 20
similarly for remaining 4 prime factors its 6, 6, 2, 2 ways
So, 20*6*6*2*2 positive integral solutions
Now, w, x, y can be positive or negative (as it doesn't matter), so 2*2*2 = 8 ways
For u and v, either both positive or both negative, so 2 ways
So, (20*6*6*2*2)*(8*2) integral solutions
Q: Find the last two digits of 1/(5^903)
GN :
GN :
@viewpt said:Q: Find the last two digits of 1/(5^903)GN :
1/5^903=2^903/10^903
now last two digits of 2^903 are 08
so ans=08
How many common roots do 2 equation have
x^3-7x^2+6x+1=0 & x^3-6x^2+x+7=0
1.0
2.3
3.2
4.4
5.none
@Logrhythm said:e(83) is 82...71^73^79%82 e(82) is 40..73^79%40 e(40) is 16..79%16 = 15=> 33^15%40 = 17...=> 11^17%82 = 69..=> 67^69%83 = 43??kya bekaar ques tha ye..
@viewpt whats the OA?
@bs0409 said:1/5^903=2^903/10^903now last two digits of 2^903 are 08so ans=08How many common roots do 2 equation havex^3-7x^2+6x+1=0 & x^3-6x^2+x+7=01.02.33.24.45.none
3.2
@bs0409 said:1/5^903=2^903/10^903now last two digits of 2^903 are 08so ans=08How many common roots do 2 equation havex^3-7x^2+6x+1=0 & x^3-6x^2+x+7=01.02.33.24.45.none
2?
subtracted both equations.. got x=2 or x=3.
subtracted both equations.. got x=2 or x=3.
@joyjitpal said:3.2
@ChirpiBird said:2?subtracted both equations.. got x=2 or x=3.
OA-0
Mr. X leaves his house daily at a particular time to go to his office. He reaches office 10 minutes late when he travels at 20 kmph and 5 minutes early when he travels at 30kmph. At what speed should he travel to reach office on time.
@bs0409 said:OA-0Mr. X leaves his house daily at a particular time to go to his office. He reaches office 10 minutes late when he travels at 20 kmph and 5 minutes early when he travels at 30kmph. At what speed should he travel to reach office on time.
D/20 = t + 10/60
D/30 = t - 5/60
solve to get t=35 and D=15
so he has to reach office in 35 mins or 35/60 hrs..
=> 15/(35/60) = 180/7 kmph...
@viewpt said:Q: Find the last two digits of 1/(5^903)GN :
08 ?
1/5^903 = 2^903/10^903 = 2^903 = 2^3*(2^10)^9 = 8*(24)^9 = 8*76 = 08